For ΔS°, keep every coefficient
Multiply each standard molar entropy by its coefficient in the balanced equation, add the products, then subtract the reactants.
ΔS° = ΣS°(products) − ΣS°(reactants)
Year 13 · AQA & OCR A
Entropy & Gibbs free energy
Practise the calculations that usually cost marks: products minus reactants for ΔS°, matching J with kJ in ΔG = ΔH − TΔS, and deciding what the sign of ΔG actually tells you.
Work out what the data are asking for
0first-time correct0completed
Before you calculate ΔG
ΔH is usually in kJ mol−1. ΔS is usually in J K−1 mol−1. Divide ΔS by 1000 before using it with ΔH in kJ mol−1.
ΔG = ΔH − TΔS
Negative ΔG means thermodynamically feasible
A negative ΔG predicts that the forward change is thermodynamically feasible at the stated temperature. ΔG = 0 is the boundary. A positive ΔG means the forward change is not thermodynamically feasible under those conditions.
OCR A: feasibility does not tell you the rate. A reaction with a negative ΔG can still be extremely slow if its activation energy is large.